Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
const double eps = 1e-10, inf = 1e9;
double s, a, b, x, y, L[maxn], R[maxn], f[maxn], ans;
int main() {
  int n, q;
  scanf("%lf%lf%lf", &s, &a, &b);
  cin >> n;
  L[1] = R[1] = 0, ++n;
  for (register int i = 2; i <= n; ++i)
    scanf("%lf%lf", &L[i], &R[i]), f[i] = f[i - 1] + R[i] - L[i];
  ++n;
  L[n] = R[n] = inf;
  f[n] = f[n - 1];
  cin >> q;
  while (q--) {
    scanf("%lf%lf", &x, &y);
    double u = (a - x) / (y - s) * y + x, v = (b - x) / (y - s) * y + x;
    if (v <= L[1] || u >= R[n])
      printf("0.000000000\n");
    else {
      int l = 0, r = n;
      while (l < r) {
        int mid = (l + r + 1) >> 1;
        if (L[mid] <= u)
          l = mid;
        else
          r = mid - 1;
      }
      ans = max(0.0, R[l] - u) - f[l];
      l = 0, r = n;
      while (l < r) {
        int mid = (l + r) >> 1;
        if (R[mid] >= v)
          r = mid;
        else
          l = mid + 1;
      }
      ans += f[r];
      ans -= min(R[l] - v, R[l] - L[l]);
      ans *= (b - a) / (v - u);
      printf("%.7lf\n", ans);
    }
  }
  return 0;
}